If the mean of the set of numbers x1,x2,…,xn is x¯, then the mean of the numbers xi+2i,1≤i≤n is

A
$\bar{x} + 2 n$
B
$\bar{x} + n + 1$
C
$\bar{x} + 2$
D
$\bar{x} + n$

Solution:

$\begin{array}{l} \bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i} \\ \Rightarrow \sum_{i = 1}^{n} x_{i} = n \bar{x} \\ Let M = \frac{1}{n} \sum_{i = 1}^{n} (x_{i} + 2 i) = \frac{1}{n} \sum_{i = 1}^{n} x_{i} + \frac{2 (1 + 2 + \dots + n)}{n} \\ \Rightarrow M = \frac{1}{n} n \bar{x} + 2 \frac{n (n + 1)}{2 n} \\ ∴ M = \bar{x} + (n + 1) \end{array}$

If the mean of the set of numbers $x_{1}, x_{2}, \dots, x_{n} is \bar{x}$ , then the mean of the numbers $x_{i} + 2 i, 1 \leq i \leq n$ is

Solution:

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